Let $R_0=\Bbb R\setminus\{0\}$. Define $*$ on $R_0$ by $a*b=\lvert a\rvert b $. Does $(R_0,*)$ form a group? 
Let $R_0$ be the set of all real numbers except zero. We define a binary operation $*$ on $R_0$ as: $a*b = \lvert a \rvert b $, where $\lvert a \rvert$ denotes absolute value of $a$. Does $(R_0,*)$ form a group?

In the above question we can see that it satisfies the closure property and associative property, but I am not sure if I am checking the existence of left identity correctly. Here is what I have solved:
Let $a \in R_0$ then $\exists e \in R_0 $ such that $e*a = a$. So from the defined operation $$e*b = \lvert e \rvert b  = b$$
$$\implies (\lvert e \rvert - 1)b =0$$
$$\implies e = 1 \text { if } e > 0 ; e = - 1 \text { if } e < 0$$
We can also check the right identity element as:
$$b*a = \lvert b \rvert e  = b$$
$$\implies e = 1 \text { if } b > 0 ; e = - 1 \text { if } b < 0$$
since the identity element inside a group is unique, so, the given binary operation is not a group.
Is my solution correct? Thanks!
 A: Suppose $G=(R_0, \ast)$ were a group. Then
$$\begin{align}
-1*2 &=|-1|2\\
&=2\\
&=|1|2\\
&=1*2
\end{align}$$
implies $-1=1$ by cancellation, a contradiction. Hence $G$ is not a group.

Alternatively, consider the following lemma:

Lemma: The only idempotent of a group is the identity $e$.

Proof: Let $x^2=x$. Then $xx=x=ex$ so, multiplying on  the right by $x^{-1}$, we get $x=e$. $\square$
But $-1\ast -1=|-1|(-1)=-1$ and $1\ast 1=|1|1=1$, while $-1\neq 1$.
Hence $G$ is not a group.
A: Almost, but not entirely. In order to search for eventual identity elements you should use equivalences (you used implications) and quantifiers:\begin{align}e\text{ is a left identity}&\iff(\forall b\in\Bbb R\setminus\{0\}):e*b=b\\&\iff(\forall b\in\Bbb R\setminus\{0\}):|e|b=b\\&\iff(\forall b\in\Bbb R\setminus\{0\}):(|e|-1)b=0\\&\iff|e|=1\\&\iff e=\pm1.\end{align}And so, as you wrote, since we got mora than one left identity, it is not a group.
A: Here is another proof:
If $x$ were the right inverse of $-5$, then it must satisfy $-5*x = |-5|x = 1$, which gives $x=\frac15$.
However, this $x$ is not the left inverse: $x*-5=\frac15\cdot-5=-1$.
So $-5$ does not have an inverse element, and $G$ is not a group.
